Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long md = 1e9 + 7;
const int xn = 1e5 + 10;
const int xm = 1e1 + 10;
const int xk = 3e1 + 10;
const int SQ = 320;
const int sq = 5e2 + 10;
const long long inf = 1e18 + 10;
int n, st[xn], ft[xn], s, t, h[xn], m, Par[xn][xk], q, c[xn];
vector<int> adj[xn], Flor[xn], dp[xn][xk], vec, ans, vis[xn];
pair<int, int> edge[xn];
bool mark[xn];
void DFS(int v) {
  mark[v] = true;
  st[v] = s++;
  Flor[h[v]].push_back(v);
  for (int i : vis[v]) dp[v][0].push_back(i);
  sort((dp[v][0]).begin(), (dp[v][0]).end());
  dp[v][0].resize(min(int(dp[v][0].size()), 10));
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (mark[u]) continue;
    h[u] = h[v] + 1;
    Par[u][0] = v;
    DFS(u);
  }
  ft[v] = s++;
}
int find_par(int v, int d) {
  int l = 0, r = Flor[d].size(), mid;
  while (r - l > 1) {
    mid = (l + r) / 2;
    int u = Flor[d][mid];
    if (st[u] <= st[v])
      l = mid;
    else
      r = mid;
  }
  return Flor[d][l];
}
bool check(int v, int u, int d) {
  int p = find_par(v, d);
  return (st[p] <= st[u] && ft[p] >= ft[u]);
}
int LCA(int v, int u) {
  if (h[v] > h[u]) swap(v, u);
  int l = 0, r = h[v] + 1, mid;
  while (r - l > 1) {
    mid = (l + r) / 2;
    if (check(v, u, mid))
      l = mid;
    else
      r = mid;
  }
  return find_par(v, l);
}
void merge(int i, int j, int k) {
  int p1 = 0, p2 = 0;
  vec.resize(0);
  while (p1 < dp[i][k].size() || p2 < dp[j][k].size()) {
    if (p1 == dp[i][k].size() ||
        (p2 < dp[j][k].size() && dp[j][k][p2] < dp[i][k][p1])) {
      vec.push_back(dp[j][k][p2]);
      p2++;
      continue;
    } else {
      vec.push_back(dp[i][k][p1]);
      p1++;
      continue;
    }
  }
  vec.resize(min(10, int(vec.size())));
}
void merg(int i, int k) {
  int p1 = 0, p2 = 0;
  vec.resize(0);
  while (p1 < dp[i][k].size() || p2 < ans.size()) {
    if (p1 == dp[i][k].size() || (p2 < ans.size() && ans[p2] < dp[i][k][p1])) {
      vec.push_back(ans[p2]);
      p2++;
      continue;
    } else {
      vec.push_back(dp[i][k][p1]);
      p1++;
      continue;
    }
  }
  vec.resize(min(10, int(vec.size())));
  ans.resize(0);
  for (int i = 0; i < vec.size(); i++) ans.push_back(vec[i]);
}
void Solve(int v, int u, int t) {
  int pnt = 30;
  ans.resize(0);
  int par = LCA(v, u);
  while (v != par) {
    while ((1 << pnt) > h[v] - h[par]) pnt--;
    merg(v, pnt);
    v = Par[v][pnt];
  }
  pnt = 30;
  while (u != par) {
    while (h[u] - pow(2, pnt) < h[par]) pnt--;
    merg(u, pnt);
    u = Par[u][pnt];
  }
  merg(par, 0);
  cout << min(t, int(ans.size())) << ' ';
  for (int i = 0; i < min(int(ans.size()), t); i++) cout << ans[i] << ' ';
  cout << '\n';
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    adj[v].push_back(u), adj[u].push_back(v);
  }
  for (int i = 0; i < m; i++) cin >> c[i], vis[c[i]].push_back(i + 1);
  h[1] = 1;
  DFS(1);
  for (int j = 1; j <= 30; j++) {
    for (int i = 1; i <= n; i++) {
      if (h[i] < pow(2, j)) continue;
      Par[i][j] = Par[Par[i][j - 1]][j - 1];
    }
  }
  for (int j = 1; j <= 30; j++) {
    for (int i = 1; i <= n; i++) {
      if (h[i] < pow(2, j)) continue;
      int p = Par[i][j - 1];
      merge(i, p, j - 1);
      for (int k = 0; k < min(10, int(vec.size())); k++)
        dp[i][j].push_back(vec[k]);
    }
  }
  while (q--) {
    int v, u, t;
    cin >> v >> u >> t;
    Solve(v, u, t);
  }
  return 0;
}